How much NaC3H5O3 (in grams) is needed to create a 1000 ml volume of Ringer’s lactate solution?
To determine the amount of NaC3H5O3 (sodium lactate) needed to create a 1000 ml volume of Ringer's lactate solution, we need to know the concentration of NaC3H5O3 in the solution.
Ringer's lactate solution typically contains 130-145 mEq/L (milliequivalents per liter) of sodium lactate. Assuming a concentration of 140 mEq/L, we can calculate the amount of NaC3H5O3 needed.
The molar mass of NaC3H5O3 is calculated as follows:
Na (sodium) - 22.99 g/mol
C (carbon) - 12.01 g/mol x 3 = 36.03 g/mol
H (hydrogen) - 1.01 g/mol x 5 = 5.05 g/mol
O (oxygen) - 16.00 g/mol x 3 = 48.00 g/mol
Adding up all the atomic masses gives us:
22.99 + 36.03 + 5.05 + 48.00 = 112.07 g/mol
In one mole of NaC3H5O3, there is one mole of Na. Thus, the molar mass of NaC3H5O3 is equivalent to its molecular weight.
To calculate the amount of NaC3H5O3 needed in grams, we can use the following formula:
Mass (in grams) = (Volume x Concentration x Molecular Weight) / 1000
Substituting the values:
Mass (in grams) = (1000 ml x 140 mEq/L x 112.07 g/mol) / 1000
Calculating this expression will give you the amount of NaC3H5O3 in grams needed to create a 1000 ml volume of Ringer's lactate solution.